An infinite cyclic group has two generators. Does the cardinality of the infinite set matter?

Question

I was reading in a book that an infinite cyclic group has exactly 2 generators. Now my question is, does the cardinality of the infinite set matter? If the set of the group is the natural numbers, or if the set of the group is the real numbers, (so they have different cardinalities), do both groups have 2 generators each?

Answer 1

Neither the natural numbers nor the real numbers are cyclic groups. Up to isomorphism, there is only one infinite cyclic group, $\mathbb{Z}$, which has two possible choices of generator: $1$ and $-1$.

Answer 2

Try to generate the real numbers as an additive group. If use a finite list, how big of a subgroup of $\mathbb{R}$ do you get? You will fail to get all of $\mathbb{R}$. For $\mathbb{Q}$, take the least common denominator $d$ of all the fractions you used in that list of generators, can you get $\frac{1}{3d}$ with those generators and just using addition and subtraction?

Answer 3

Every infinite cyclic group is isomorphic to $\mathbb{Z}$ and $\mathbb{Z}$ is generated by two elements which are $1$ and $-1$.

$\mathbb{R}$ is not a cyclic groue.

Answer 4

To clarify an ambiguity a definition might help: a group $G$ is defined to be cyclic if it has a generating set consisting of a single element $g$; and any such element $g$ is called a generator of $G$.

By combining this definition with the definition of a generating set, one obtains a surjective function $\mathbb{Z} \mapsto G$ defined by $n \mapsto g^n$.

One can then go on to prove the fact mentioned in other answers, saying that every countably infinite cyclic group $G$ is isomorphic to $\mathbb{Z}$. This follows from the fact that the function $n \mapsto g^n$ is a homomorphism, that it is surjective, and that its kernel is trivial (because otherwise its image is finite).